The correct order of increasing s character (in percentage) in the hybrid orbitals in below ions is (assume all hybrid orbitals are exactly equivalent) :

$C O_{3}^{2 -} I, X e F_{4} I I, I_{3}^{-} I I I, N C l_{3} I V, B e C l_{2} V$

I I < I I I < I V < I < V

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I I < I V < I I I < V < I

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I I I < I I < I < V < I V

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I I < I V < I I I < I < V

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Solution

The correct option is A $I I < I I I < I V < I < V$
$B e C l_{2} \to s p \to 50 %$ maximum

$X e F_{4} \to s p^{3} d^{2} \to 10.66 %$ (minimum)

$N C l_{3} \to s p^{3} \to 25 %$

$I_{3}^{-} \to s p^{3} d \to 20 %$

The correct order would be:
II<III<IV<I<V

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Similar questions

\begin{matrix} C O_{3}^{2 -} & X e F_{4} & I_{3}^{-} & N C l_{3} & B e C l_{2} (I) & (I I) & (I I I) & (I V) & (V) \end{matrix}

C O_{3}^{2 -} I; X e F_{4} I I; I_{3}^{-} I I I; N C l_{3} I V; B e C l_{2} (g) V

s -

C O_{3}^{2 -}

X e F_{4}

I_{3}^{-}

N C l_{3}

B e C l_{2}

: C H_{2} X e F_{4} B r F_{3} N C l_{3} : C C l_{2}

\begin{matrix} C O_{3}^{2 -} & X e F_{4} & I_{3}^{-} & N C l_{3} & B e C l_{2} (I) & (I I) & (I I I) & (I V) & (V) \end{matrix}

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